//验证尼科彻斯定理，即：任何一个整数 m 的立方都可以写成 m 个连续奇数之和
//1^3=1               
//2^3=3+5 
//3^3=7+9+11
//4^3=13+15+17+19
//2^(n-1)+1   1  2/3   4/5/6   7/8/9/10 
#pragma warning(disable:4996)
#include<stdio.h>
int Odd_Sum(int n)
{
	int sum = 1;
	for (int i = 1; i < n; i++) {
		sum += 2;
	}
	return sum;
}
int main()
{
	int m, start, first, times;
	
	while (~scanf("%d", &m))
	{
		start = 0;
		for (int i = 1; i < m; i++)
		{
			start += i;
		}
		first = Odd_Sum(start) + 2;
		times = m;
		while (times) {
			if (times == 1) {
				printf("%d", first);
				break;
			}
			printf("%d+", first);
			first += 2;
			times--;
		}
	}
	return 0;
}